Chapter 7, Problem 7.59FU

Interpretation Introduction

(a)

Interpretation:

The molar mass of clavulanic acid should be determined.

Concept Introduction:

Molar mass of compound is equal to the sum of the atomic masses of each element that makes up the compound.

Answer to Problem 7.59FU

The molar mass of clavulanic acid = 199 g/mol

Explanation of Solution

Atomic mass of C = 12 g/mol

Atomic mass of H = 1 g/mol

Atomic mass of N = 7 g/mol

Atomic mass of O = 16 g/mol

The molar mass of clavulanic acid, ${\text{C}}_{\text{8}}{\text{H}}_{\text{9}}{\text{NO}}_{\text{5}}$ = $\left(\text{12 }\times \text{ 8}\right)\text{ + }\left(\text{1 }\times \text{ 9}\right)\text{ + }\left(\text{14 }\times \text{ 1}\right)\text{ + }\left(\text{16 }\times \text{ 5}\right)\text{ = 199 g/mol}$

Interpretation Introduction

(b)

Interpretation:

The mass percent of C in clavulanic acid should be determined.

Concept Introduction:

  $\text{Mass percent =}\frac{\text{mass of element in the compound}}{\text{molar mass of the compound}}\text{×}\text{ 100}$

Answer to Problem 7.59FU

Mass percent of C in clavulanic acid is 48.24 %.

Explanation of Solution

Atomic mass of C = 12 g/mol

The number of C atoms in clavulanic acid is 8.

Mass percent of C in clavulanic acid:

  $\begin{array}{l}\text{= (}\frac{\text{mass of C in clavulanic acid}}{\text{molar mass clavulanic acid}}\text{) }\times \text{ 100}\\ \text{= }\frac{\left(\text{12}\times \text{8}\right)}{\text{199}}\times \text{100 = 48}\text{.24 \%}\end{array}$

Interpretation Introduction

(c)

Interpretation:

Moles of clavulanic acid given should be determined.

Concept Introduction:

Relation between mass and molar mass:

  $\text{mole}\text{(mol)}\text{=}\frac{\text{mass(g)}}{\text{molar}\text{mass(g/mol)}}$

Answer to Problem 7.59FU

Moles of clavulanic acid given to Max is $\text{3}\text{.6432 }\times {\text{ 10}}^{\text{-4}}\text{mol}$ .

Explanation of Solution

Weight of Max = 29 kg

Dose of clavulanic acid = 2.5 mg/kg

Mass of clavulanic acid given to Max:

  $\begin{array}{l}\text{=}\text{29 kg }\times \text{ 2}\text{.5 mg/kg}\\ \text{= 72}\text{.5 mg = 72}\text{.5 }\times {\text{10}}^{\text{-3}}\text{g}\end{array}$

Moles of clavulanic acid given to Max:

  $\begin{array}{l}\text{= }\frac{\text{mass of clavulanic acid}}{\text{molar mass clavulanic acid}}\\ \begin{array}{l}\begin{array}{l}\\ \text{= }\frac{\text{72}\text{.5 }\times {\text{ 10}}^{\text{-3}}\text{g}}{\text{199 g/mol}}\end{array}\hfill \\ \text{= 3}\text{.6432 }\times {\text{ 10}}^{\text{-4}}\text{mol}\hfill \end{array}\end{array}$